Chapter 4 Bipolar Junction Transistors
Outline
 Basic operation of the npn Bipolar Junction Transistor
 Load-line analysis of a common-emitter amplifier
 The pnp bipolar junction transistor
 Small-signal equivalent circuits
 The common-emitter amplifier
 The emitter follower (common-collector amplifier)
 The common-base amplifier
4.1 Basic operation of the npn transistor
 A npn BJT consists of a thin layer of p-type
material between two layers of n-type
material.
 Two pn junctions are formed in the device.
 The current flowing across one junction
affects the current in the other junction.
It is this interaction that makes the BJT
useful as an amplifier.
Figure 4.1 The npn BJT
4.1 Basic operation of the npn transistor
 Basic operation in the active region
 The common-emitter configuration
 Active region (normal operation)
 Base-emitter junction is forward biased
 The base-collector junction is reverse biased
The emitter region is doped heavily.
The base region is very thin.
Figure 4.2 An npn transistor with variable
biasing sources (CE configuration)
4.1 Basic operation of the npn transistor
 Amplification by the BJT
Figure4.4 CE characteristics of a typical npn BJT
A small change in vbe can result in an appreciable change in ib, this causes a
much larger change in ic, and in suitable circuits, it is converted into a much
larger voltage change than the initial change in vbe.
4.1 Basic operation of the npn transistor
 The common-emitter current gain
iC

iB
 Device equations
10~1000
iE  iB  iC
 The common-base current gain
iC

iE
Exercise:
1.
Find the relationship between  and 
2.
Estimate  and  for Figure4.4
4.2 Load-line analysis of a CE amplifier
The power-supply voltages VBB and VCC
bias the device at an Q point for which the
amplification of the input signal is possible.
Exercise:
1. Write KVL for input loop.
2. Write KVL for output loop.
P144 Example 4.2
Figure4.6 Common emitter amplifier
4.2 Load-line analysis of a CE amplifier
Pay attention to
Phase relationship
between vin and vout
4.2 Load-line analysis of a CE amplifier (ref 4.4)
 Active region model — amplification (1)
 Saturation region model (2)
 Cutoff region model (3)
 Inverted (reverse) model (4)
Model
B-E junction
B-C junction
(1)
forward
reverse
(2)
forward
forward
(3)
reverse
reverse
(4)
reverse
forward
Figure 4.11 Amplification in the active region
4.2 Load-line analysis of a CE amplifier
Distortion occurs?
Figure Distortion illustration
4.3 The pnp bipolar junction transistor
Example: Determine the diode states for the circuits shown below.
Assume ideal diodes.
Think about:
1. Circuit configuration
Figure 4.12 The pnp BJT
2. Active region — source polarity
4.4 Large-signal DC analysis of BJT circuit
Example: Find the Q-point (VBEQ, IBQ, VCEQ, ICQ) of the circuit (P151)
Think about:


1.
Indicate the current flowing
through RB and RC
=100
2.
Write KVL for input/output loop
Figure 4.13 The BJT
VBEQ is given (about 0.7V for Si, and 0.2V for Ge)
IBQ=(VCC-VBEQ)/RB
ICQ= IBQ
VCEQ=VCC-ICQRC
4.4 Large-signal DC analysis of BJT circuit
Example: Find the Q-point of the circuit (P174)
Think about:
1.
Indicate the current flowing
through RB，RE and RC
2.
Write KVL for input/output loop
IBQ=(VB-VBEQ)/[RB+(1+)RE]
ICQ= IBQ
Figure
3.11 Half-wave rectifier with resistive load
Figure 4.14 BJT with two voltage
supplies
VCEQ=VCC-IEQRE-ICQRC VCC-ICQ(RC+RE)
4.4 Large-signal DC analysis of BJT circuit
Analysis of the four-resistor bias circuit
Method 1 — Thevenin equivalent circuit
Convert to Figure4.14
+
VBQ
-
VCEQ
Method 2 — Assume IR1>>IBQ, IR2>>IBQ
R1 and R2 are in series
VBQR2VCC/(R1+R2)
Figure 4.15 The BJT with four resistors
IEQ=(VBQ-VBEQ)/RE
VCEQ=VCC-IEQRE-ICQRC VCC-ICQ(RC+RE)
4.5 Small-signal equivalent circuit
Input resistance rbe
rce
rbe(r)=300+(1+)26(mV)/IEQ(mA)
Tansconductance gm
gm= /rbe
Output resistance rce
Usually large, negligible
rce
Figure 4.16 small signal equivalent circuits for the BJT
4.6 The common-emitter amplifier (P164)
 DC supply voltage —— bias the device
at a suitable operating point
 AC signal —— what we want to amplify
 The use of C1, C2 and Ce
C2
 Coupling — C1, C2
Ce
 bypass — Ce
 Circuit configuration
Figure 4.28 Common-emitter amplifier of Example 4.9.
 What shall we do with this circuit?
 DC bias circuit — determine Q point
 Small signal equivalent circuit
 Find Av, Rin and Rout
4.6 The common-emitter amplifier (P164)
1.
Draw the DC bias circuit to find the Q point

Ac signal — short circuit

Capacitor — open circuit
+
VBQ
Figure 4.28 Common-emitter amplifier of Example 4.9.
Rule of thumb: VCEQ 0.3~0.7VCC
-
VCEQ
4.6 The common-emitter amplifier (P164)
2. Draw the AC circuit with BJT

DC signal — Connect to the GND

Capacitor — short circuit
3. Replace with the equivalent circuit
Figure 4.28 Common-emitter amplifier of Example 4.9.
rce
4.6 The common-emitter amplifier (P164)
rce
Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0
Think about: 1. the relationship between R1, R2 and rce
2. the relationship between rce, RC and RL
3. Write two equations for Vin and Vo
4.6 The common-emitter amplifier (P164)
Find the voltage gain Av=Vo/Vin
Find the open circuit voltage gain Avo=Vo/Vin|RL=0
Find the source voltage gain Avs=Vo/Vs
Find the input resistance Rin=Vin/Iin
Find the output resistance Rout=Vo/Io|Vs=0
rce
Rout
4.6 The common-emitter amplifier (P164)
Analysis of the CE amplifier without CE
DC operating point
C2
Remains unchanged
Ce
AC performance
Need more detailed analysis
Exercise: 1. Draw the small-signal equivalent circuit
2. Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0 assume rce
3. Compare with the results for circuit with CE
4.6 The common-emitter amplifier (P164)
Think about: 1. the currents flowing through rbe, RE and RL’
2. Write two equations for Vin and Vo
4.6 The common-emitter amplifier (P162)
Figure 4.27 Common-emitter amplifier.
Think about: 1. Effects of RE1 on Q-point, Av,Rin and Rout
2. Effects of RE2 on Q-point, Av,Rin and Rout
4.7 The emitter follower (P166)
1.
Draw the DC bias circuit to find the Q point
Given R1=R2=100k, =200, VBEQ=0.7V
find the Q point.
4.7 The emitter follower (P166)
R1
RL
(b) AC circuit
4.7 The emitter follower (P166)
Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0
Think about: Write two equations for Vin and Vo
RS’
Rout’
Rout
4.8 The common-base amplifier (P299)
1.
Draw the DC bias circuit to find the Q point
4.8 The common-base amplifier (P299)
Find Av=Vo/Vin, Rin=Vin/Iin and Rout=Vo/Io|Vs=0
Think about: Write two equations for Vin and Vo
RC
Rin
Rout’
Rout
4.8 The common-base amplifier (P299)
 Summary
C-E amplifier
C-C amplifier
C-B amplifier
inverting
noninverting
noninverting
amplification
follower
amplification
Av
Ri
Ro
Review 1.5 Cascaded amplifiers
 The overall voltage gain of cascaded amplifier stages is
the product of the voltage gains of the individual stages.
 Think over: If Avo1=100, Avo2=200, what is the overall
open circuit voltage gain of cascaded amplifier?

Avo=Avo1 Avo2 ?
(refer to p17)
Review 1.5 Cascaded amplifiers
 Applications calling for high or low input impedance
 Applications calling for high or low output impedance
 Application calling for a particular impedance
 Refer to examples shown on page 26-27
4.9 Cascode amplifier (P300)
4.9 Cascode amplifier (P300)
4.9 Cascode amplifier (P300)
 共集－共射组合放大电路，不仅具有共集电极电路输入电阻大的
特点，而且具有共射电路电压放大倍数大的特点；
 共射－共集组合放大电路，不仅具有共射电路电压放大倍数大的
特点，而且具有共集电极电路输出电阻小的特点；
 共射－共基组合放大电路，共基极电路本身就有较好的高频特性，
同时将输入电阻很小的共基极电路接在共射极电路之后，减小了
共射极电路的电压放大倍数，使共射极接法的管子集电结电容效
应减小，改善了放大电路的频率特性。因此，共射－共基组合放
大电路在高频电路中获得了广泛的应用。该组合电路的电压放大
倍数近似等于一般共射电路的电压放大倍数。
AC coupling versus direct coupling (P32)
Figure 1.37 Capacitive coupling illustration
AC coupling — The DC voltages of the amplifier circuits do not affect the
signal source, adjacent stages, or the load.
Amplifiers that are realized as integrated circuits are almost always DC
coupled because the capacitors or transformers needed for ac coupling
cannot be fabricated in integrated form.
BACK
Example
4.10 Frequency response for an amplifier
 Frequency response (P30)
 The complex gain: The ratio of the phasor for the output signal to
the input signal
 Bode plot (P271)
 How circuit functions can be quickly and easily plotted against
frequency? (straight line approximation & smart scale)
Review
Logarithmic Frequency Scale
A decade is a range of frequencies for which the ratio
of the highest frequency to the lowest is 10.
An octave is a two-to-one change in frequency.
Review: Passive low-pass filter
Figure 8.1 Low-pass RC filter.
Review 2.8 Active Filter-High pass filter
=
4.10 Frequency response for an amplifier
Question:
1. What is the voltage
gain of the circuit?
2. Can you draw the
frequency response of
amplifier?
3. What kind of filter is it?
4. Do you remember the
small signal equivalent
circuit of the NPN
transistor?
5. Any changes to (4) if
the frequency is high?
4.10 Frequency response for an amplifier
Cbc
Cbe
Mid-frequency
The small signal equivalent
circuit for the NPN transistor
At high frequencies
Usually, Cbe is about 10PF ~ 1000PF
Cbc <10PF
Miller effect
Question: (P296)
If Zf=-j/C for a CE
amplifier, how do
you find Zin,miller
and Zout,miller
Find the Thevenin equivalent resistance Rs’?
(1+gmRL’)Cbc
Cbc
rbe
Cbe
 ib
4.10 Frequency response for an amplifier
Figure 8.34 Simplified equivalent circuit for the common-emitter amplifier.
1. What kind of filter is it?
2. What is the break (cut-off) frequency?
3. What factors affect the value of the break frequency?
4. What do we desire about an amplifier?
4.10 Frequency response for an amplifier
Figure 8.36 High-frequency behavior of the common-emitter amplifier.
Conclusion: The common-base amplifier achieves wide
bandwidth, but its input impedance is very low.
Consequently, the mid-band gain can be quite small due
to loading of source. CE-CC cascade amplifier combines
the advantages of both.
4.10 Frequency response for an amplifier
Figure 8.47 Amplifier with coupling capacitors.
Figure 8.47 Amplifier with coupling capacitors.